``passes through `(-4, 1)`, perpendicular to a line whose slope is `-3/2` Graph the line that satisfies each set of conditions.

Textbook Question

Chapter 2, 2.3 - Problem 44 - Glencoe Algebra 2 (1st Edition, McGraw-Hill Education).
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kspcr111 | In Training Educator

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The line which we needed is perpendicular to a line whose slope is `(-3/2)`  .

the product of the slopes of two lines which are perpendicular is equal to `-1`

let the slope of the line which we need to find be `m_1` and the slope of the other line  be `m_2 = (-3/2).`

so ,

`m_1 * m_2 = -1`

=> `m_1 = -1/(m_2) = -1 /(-3/2) = 1/(3/2) = 2/3`

so , `m_1 = 2/3`

and the line of slope `m_1` passes through the point `(-4,1)`

then the line is 

`y = mx+c`

=> `y = (2/3)x +c`

as it passes through `(x,y)=(-4,1)` so 

=>  `1= (2/3)(-4) +c`

=>` 1= -8/3 +c`

=> `1+8/3 = c`

=> `c = 11/3`

so the equation of the line is `y= (2/3)x+ 11/3` and the graph plotted is as follows in the attachments. the point `(-4,1)` is spotted with a green dot.

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